I have a game where you start with 13 cards (for example a single suit). Randomly select three cards, from which you discard both the lowest and the highest, while returning the median card to the original set. This process is then repeated with the sucessively smaller set until only a single card remains. What is the probability for each of the 13 cards you started with remaining as the last one in the set?

N=1 and N=13 have zero odds of winning the game. Since N=2 has 1 card of lower value, and N=12 has 1 card of higher value, they have the same odds of winning the game. Similarly N=3 and N=11 have the same odds, N=4 and N=10 have the same odds, N=5 and N=9 have the same odds, and N=6 and N=8 have the same odds.

With a deck of 13 cards, with 3 cards drawn at a time, the larger and smaller card discarded, and the middle one returned to the deck, there will be six drawings. In round 1, there are 13 cards, and each card has 3/13 odds of being selected. In round 2, 11 cards with 3/11 odds per card of being selected. In round 3, 9 cards with 1/3 odds per card. In round 4, 7 cards with 3/7 odds per card. In round 4, 5 cards with 3/5 odds. In round 6, 3 cards are left; two will be eliminated, leaving the "winning card".

In rounds 1-5, the odds of card N surviving the round, are the odds that it survived the prior round (or 1 for round 1) times the odds that it will survive this round.

The odds that it will survive this round are the odds that it will not be selected, plus the odds that both it and exactly one smaller card will be selected.

The odds that exactly one smaller card will be selected, is the sum of the odds that each smaller card was the only smaller card selected. That is the odds that the smaller card survived the prior round (1 for round 1) times the odds that the smaller card will be selected this round, times the product, for every other smaller card, of the complement of the odds that it survived the prior round (0 for round 1) plus the odds that it did survive the prior round times the complement of the odds that it will be selected in this round.

In round 6, the odds for the survival of N=1 and N=13 are zero. For N=2,3,...,12, the odds are the odds that N survived round 5 and exactly one smaller card survived round 5. The latter is the sum, for each smaller card, of the odds that it survived round 5, times the product, for all other smaller cards, of the complement of the odds that it survived round 5.

Now I was morally certain that the above reasoning produces the correct results. At this point I had the option of tediously computing everything by hand, with the risk of cascading errors, or shamelessly borrowing from Charlie. (I will have to learn how to convert logic into programs as Charlie does. Maybe that will be my 2004 New Years Resolution).

The results are:

1 has 0.0000000 odds of being the winning card
2 0.0034965
3 0.0127146
4 0.0357067
5 0.0892845
6 0.1961927
7 0.3252101
8 0.1961927
9 0.0892845
10 0.0357067
11 0.0127146
12 0.0034965
13 0.0000000
Edited on January 4, 2004, 2:51 pm

Posted by Penny on 2004-01-03 18:04:42